Question

In Problem, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers n.

$n^2-n+2$is divisible by$2$

Short answer

The given statement is shown.

Step-by-step solution

Step 1. Given information

The statement is$n^2-n+2$

Step 2. Show that the given statement is true.

Put $n=1$in $n^2-n+2$to check if the statement is true for $n=1$.

width="149">(1)2-1+2=1-1+2=2

2 is divisible by 2.

Thus, the statement is true for width="37">n=1.

Assume that the statement is true for $n=k$.

Thus, $k^2-k+2$is divisible by 2

Find whether $(k+1{)}^2-(k+1)+2$is divisible by 2.

$\begin{array}{rcl}(k+1{)}^2-(k+1)+2& =& k^2+2k+1-k-1+2\\ & =& k^2-k+2+2k\\ & =& \left(k^2-k+2\right)+2k\end{array}$

Since $k^2-k+2$is divisible by 2 and $2k$is also divisible by $2,(k+1{)}^2-(k+1)+2$is divisible by 2 . Thus, the statement is true for width="34">k+1

Since both the conditions of the Principle of Mathematical Induction are satisfied, the statement is true for all natural numbers.

Therefore, $n^2-n+2$is divisible by 2